How do you solve the equation by factoring and the zero-product rule: $4 x^{2} - 25$ $4x^2-25$ ?

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Answer 1 · 2016-12-06T18:26:51

$\pm \frac{5}{2}$ $+-5/2$

Method 1

$4 x^{2} - 25 = 0$ $4x^2-25=0$

factorise by difference of squares

$(2 x + 5) (2 x - 5) = 0$ $(2x+5)(2x-5)=0$

solving for each bracket

$2 x + 5 = 0 \Rightarrow 2 x = - 5 \Rightarrow x = - \frac{5}{2}$ $2x+5=0=>2x=-5=>x=-5/2$

$2 x 5 = 0 \Rightarrow 2 x = 5 \Rightarrow x = \frac{5}{2}$ $2x5=0=>2x=5=>x=5/2$

Method 2

rearrange the eqn for $x$ $""x$

$4 x^{2} - 25 = 0 \Rightarrow 4 x^{2} = 25$ $4x^2-25=0=>4x^2=25$

$x^{2} = \frac{25}{4}$ $x^2=25/4$

$x = \pm \sqrt{\frac{25}{4}} = \pm \frac{5}{2}$ $x=+-sqrt(25/4)=+-5/2$

How do you solve the equation by factoring and the zero-product rule: 4x^2-254x2−254x^2-25?